Question

An electron moves in a circular orbit with a uniform speed $v$. It produces a magnetic field $B$ at the centre of the circle. The radius of the circle is proportional to

• A. $\sqrt{\frac{B}{v}}$
• B. $\frac{B}{v}$
• C. $\sqrt{\frac{v}{B}}$
• D. $\frac{v}{B}$

Answer 1

The correct option is C $\sqrt{\frac{v}{B}}$

Applying Biot- Savart law on moving charge $q$ with velocity $\overrightarrow{v}$

$\overrightarrow{B}=\frac{{\mu }_0}{4\pi }\frac{q(\overrightarrow{v}\times \overrightarrow{r})}{r^3}$

Here, $r$ becomes radius of circular path in given problem as magnetic field $B$ is calculated at centre.


From the diagram, we can see that $\overrightarrow{v}$ is always perpendicular to $\overrightarrow{r}$ in circle.

$\therefore B=\frac{{\mu }_{0}qvr\sin {90}^{\circ }}{4\pi r^3}=\frac{{\mu }_{0}qv}{4\pi r^2}$

$\Rightarrow r=\sqrt{\frac{{\mu }_{0}qv}{4\pi B}}$

For a given charge particle, $q=-e$,

$r\propto \sqrt{\frac{v}{B}}$

Therefore, option $\left(c\right)$ is correct.